Q.Find the equation of the set of points which are equidistant from the points (1,2,3) and (3,2,−1).
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The Locus of a Point: From Intuition to Precision
Imagine you're walking in a park, but you must always stay exactly 5 metres away from a fountain at the centre. As you walk, your path traces out a circle. That circle is the locus of your position — the set of all points that satisfy the rule "distance from fountain = 5 m".
Now think of a different rule: you must always be equally far from two trees. Your path becomes the perpendicular bisector of the line joining those trees — a straight line.
Every geometric shape you know — circle, line, parabola, ellipse — is really just a locus. A circle is the set of points at a fixed distance from a centre. A line is the set of points that satisfy a linear equation. The word "locus" (plural: loci) simply means "place" or "path" in Latin.
The Precise Definition
Locus of a point is the set of all positions (points) that satisfy a given geometric condition or a set of conditions.
In coordinate geometry, a locus is represented by an equation in x and y (or x, y, z in 3D). Every point (x,y) that satisfies the condition lies on the locus; every point that does not satisfy it lies off the locus.
How to Find the Equation of a Locus
The process is mechanical. Suppose a point P(x,y) moves so that its distance from a fixed point A(2,3) is always 5 units.
- Write the condition in words: Distance PA=5.
- Translate into algebra: (x−2)2+(y−3)2=5.
- Simplify: Square both sides: (x−2)2+(y−3)2=25.
That's it. The locus is a circle with centre (2,3) and radius 5.
A Slightly Harder Example
Find the locus of a point P(x,y) that moves so that its distance from A(1,0) is twice its distance from B(4,0).
Step 1 — Condition: PA=2⋅PB.
Step 2 — Algebra:
(x−1)2+y2=2(x−4)2+y2
Step 3 — Square and simplify:
(x−1)2+y2=4[(x−4)2+y2]
x2−2x+1+y2=4(x2−8x+16+y2)
x2−2x+1+y2=4x2−32x+64+4y2
0=3x2−30x+3y2+63
x2−10x+y2+21=0
Step 4 — Complete the square:
(x2−10x+25)+y2=4
(x−5)2+y2=4
The locus is a circle with centre (5,0) and radius 2.
When the condition involves distances in a ratio, the locus is often a circle (called the Apollonius circle). If the ratio is 1:1, the locus is the perpendicular bisector — a straight line.
Common Loci You Must Know
| Condition | Locus | Equation (standard form) |
|---|---|---|
| Fixed distance from a point | Circle | (x−h)2+(y−k)2=r2 |
| Equal distances from two points | Perpendicular bisector | Linear equation |
| Fixed distance from a line | Pair of parallel lines | $ |
| Sum of distances from two fixed points is constant | Ellipse | a2x2+b2y2=1 |
Set the squared distances from (1,2,3) and (3,2,−1) equal and simplify. …
Equating the distances from (1,2,3) and (3,2,−1) and simplifying gives the plane x−2z=0.
Let P(x,y,z) be a point equidistant from A(1,2,3) and B(3,2,−1). Then PA=PB, so PA2=PB2:
(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2
The (y−2)2 term cancels from both sides. Expand the rest:
x2−2x+1+z2−6z+9=x2−6x+9+z2+2z+1
−2x−6z+10=−6x+2z+10
−2x−6z=−6x+2z …
- CBSE 2023Set 1B4 marksQ.Find the equation of the locus of P, if the ratio of distances from P to A(5,−4) and B(7,6) is 2:3.
›Reveal solutionSolution
Set 3⋅PA=2⋅PB, square both sides, and simplify using the distance formula.
Let P=(x,y). Given PBPA=32 where A=(5,−4), B=(7,6), i.e. 3PA=2PB, so 9PA2=4PB2.
PA2=(x−5)2+(y+4)2=x2+y2−10x+8y+41
PB2=(x−7)2+(y−6)2=x2+y2−14x−12y+85
…
- CBSE 2022Set 1B4 marksQ.If the distances from P to the points (2,3) and (2,−3) are in the ratio 2:3, then find the equation of the locus of P.
›Reveal solutionSolution
Write PA:PB=2:3 as 9PA2=4PB2 using the distance formula, then simplify.
Let P=(x,y), A=(2,3), B=(2,−3). Given PBPA=32, so 3PA=2PB, i.e. 9PA2=4PB2.
PA2=(x−2)2+(y−3)2,PB2=(x−2)2+(y+3)2
So:
9[(x−2)2+(y−3)2]=4[(x−2)2+(y+3)2]
Group the (x−2)2 terms and expand the y terms:
5(x−2)2+[9(y−3)2−4(y+3)2]=0
9(y−3)2=9y2−54y+81, and 4(y+3)2=4y2+24y+36, so: …
- CBSE 2022Set 1B4 marksQ.A(5,3) and B(3,−2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9.
›Reveal solutionSolution
Use the area-of-triangle formula with P(x,y), A(5,3), B(3,−2), set it equal to 9, and solve.
Let P=(x,y), A=(5,3), B=(3,−2). The area of triangle PAB is
Area=21x(3−(−2))+5(−2−y)+3(y−3)
=215x−10−5y+3y−9=215x−2y−19
Setting the area equal to 9:
21∣5x−2y−19∣=9⟹∣5x−2y−19∣=18
…
- CBSE 2021Set annual24 marksQ.Find the equation of the set of points P such that (PA)2+(PB)2=2k2, where A and B are the points (3,4,5) and (−1,3,−7) respectively.
›Reveal solutionSolution
Expanding (PA)2+(PB)2=2k2 for A(3,4,5), B(−1,3,−7) gives 2x2+2y2+2z2−4x−14y+4z+109−2k2=0.
Let P=(x,y,z). Then:
PA2=(x−3)2+(y−4)2+(z−5)2
PB2=(x+1)2+(y−3)2+(z+7)2
Expanding each pair of terms:
(x−3)2+(x+1)2=2x2−4x+10
(y−4)2+(y−3)2=2y2−14y+25
(z−5)2+(z+7)2=2z2+4z+74
So:
PA2+PB2=2x2+2y2+2z2−4x−14y+4z+(10+25+74)
=2x2+2y2+2z2−4x−14y+4z+109
…
- CBSE 2018Set 1B4 marksQ.Find the equation of the locus of P, if A=(2,3), B=(2,−3) and PA+PB=8.
›Reveal solutionSolution
The locus is the ellipse 16(x−2)2+7y2=112 (equivalently 16x2+7y2−64x−48=0).
Let P=(x,y). Then
PA+PB=(x−2)2+(y−3)2+(x−2)2+(y+3)2=8.
Move one radical over and square:
(x−2)2+(y−3)2=64−16(x−2)2+(y+3)2+(x−2)2+(y+3)2.
The (x−2)2 cancel; (y−3)2−(y+3)2=−12y, so
−12y=64−16(x−2)2+(y+3)2⟹4(x−2)2+(y+3)2=16+3y.
Square again:
16[(x−2)2+(y+3)2]=256+96y+9y2. …
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